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Difference between revisions of "2018 AMC 10A Problems/Problem 14"

(Created page with "What is the greatest integer less than or equal to <cmath>\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?</cmath> <math> \textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \...")
 
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\textbf{(E) }625\qquad
 
\textbf{(E) }625\qquad
 
</math>
 
</math>
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 +
==Solution==
 +
 +
Let's set this value equal to <math>x</math>. We can write
 +
<cmath>\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=x.</cmath>
 +
Multiplying by <math>3^{96}+2^{96}</math> on both sides, we get
 +
<cmath>3^{100}+2^{100}=x(3^{96}+2^{96}).</cmath>
 +
Now let's take a look at the answer choices. We notice that <math>81</math>, choice <math>B</math>, can be written as 3^4. Plugging this into out equation above, we get
 +
<cmath>3^{100}+2^{100} ? 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} ? 3^{100}+3^4*2^{96}.</cmath>
 +
The right side is larger than the left side because
 +
<cmath>2^{100} \leq 2^{96}*3^4.</cmath>
 +
This means that our original value, <math>x</math>, must be less than <math>81</math>. The only answer that is less than <math>81</math> is <math>80</math> so our answer is <math>\boxed{A}</math>.

Revision as of 14:41, 8 February 2018

What is the greatest integer less than or equal to \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?\]

$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$

Solution

Let's set this value equal to $x$. We can write \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=x.\] Multiplying by $3^{96}+2^{96}$ on both sides, we get \[3^{100}+2^{100}=x(3^{96}+2^{96}).\] Now let's take a look at the answer choices. We notice that $81$, choice $B$, can be written as 3^4. Plugging this into out equation above, we get \[3^{100}+2^{100} ? 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} ? 3^{100}+3^4*2^{96}.\] The right side is larger than the left side because \[2^{100} \leq 2^{96}*3^4.\] This means that our original value, $x$, must be less than $81$. The only answer that is less than $81$ is $80$ so our answer is $\boxed{A}$.

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